Question: $\text E = \left[\begin{array}{rr}-1 & 4 \\ 5 & 2\end{array}\right]$ and $\text C = \left[\begin{array}{rr}1 & 2 \\ -1 & 1\end{array}\right]$. Let $\text {H = EC}$. Find $\text H$. $ {H = }$
Explanation: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{C}$. $ \text {H}=\left[\begin{array}{rr}{-1} & {4} \\ 5 & 2\end{array}\right]\left[\begin{array}{rr} {1} & 2 \\ {-1} & 1\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(-1,4)\cdot(1,-1)\\\\ &=-1 \cdot 1 + 4\cdot -1\\\\ &=-5 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot -1 + 2\cdot -1 = -7$ (Choice B) B $5 \cdot 1 + 2\cdot -1 = 3$ (Choice C) C $-1 \cdot 2 + 4\cdot 1 = 2$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \left[\begin{array}{rr}-5 & 2 \\ 3 & 12\end{array}\right] $